LeetCode 099 Recover Binary Search Tree

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题目

恢复BST，有两个点交换了。

代码

public class Solution {    public void recoverTree(TreeNode root) {        if(root == null){            return ;        }        ArrayList<TreeNode> record = new ArrayList<TreeNode>();        TreeNode first = null;        TreeNode second = null;        useme(record,root);                for(int i=1;i<record.size();i++){            if(record.get(i-1).val>record.get(i).val){                if(first == null){                    first = record.get(i-1);                    second = record.get(i);                }                else{                    second = record.get(i);                }            }        }        int temp = first.val;        first.val = second.val;        second.val = temp;        return;    }        public void useme(ArrayList<TreeNode> record , TreeNode root){        if(root!=null){            useme(record,root.left);            record.add(root);            useme(record,root.right);        }    }}

依然可以用inorder遍历的方法，转换为一个递增数组中，找到交换顺序的两个数。

上面的方法需要O(n)的空间，那么减少空间的方法，就是使用 Morris 中序遍历，定义再维基，百度都有，直接贴代码

public class Solution {    public void recoverTree(TreeNode root) {        if(root == null){            return ;        }        TreeNode pre = null;        TreeNode cur = root;        TreeNode first = null;        TreeNode second = null;        TreeNode a =null;        TreeNode b = null;                while(cur!=null){            if(cur.left==null){                if(first ==null) first = cur;                else if(second == null) second = cur;                else{                    first = second;                    second = cur;                }                cur = cur.right;            }            else{                pre = cur.left;                while(pre.right!=null && pre.right!=cur) pre = pre.right;                if(pre.right == null){                    pre.right = cur;                    cur = cur.left;                    continue;                }                else{                    pre.right = null;                    if(second == null) second = cur;                    else{                    first = second;                    second = cur;                    }                    cur = cur.right;                }            }            if(first!=null && second!=null && first.val > second.val){                if(a == null) a = first;                b = second;            }        }        int temp = a.val;        a.val = b.val;        b.val = temp;    }}

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